Rate and Work Problems

Intuition

The last lesson left you with a picture: put more equally fast workers on a job and it finishes sooner, drive faster and a trip takes less time. Both are inverse variation, time pulling against effort, as long as every worker shares the same steady rate. Real workers rarely match speeds exactly, so this lesson turns that picture into a method that still works when they do not, in two parts. First: work problems, where a job gets finished by one or more workers, each tracked by their own rate instead of just a headcount. Then: travel problems, where a trip covers a distance at a speed. Both parts rest on the same shift in viewpoint. Stop asking how long a job takes and start asking how fast it gets done, its rate. A painter who finishes a room in $4$ hours paints a quarter of it each hour, so the rate is $\frac{1}{4}$ of the room per hour. Rates have a property that times do not: when two workers hold steady rates and pitch in on the same job at once, without duplicating each other’s work or getting in each other’s way, their rates simply add. That single fact turns every “how long together” question into arithmetic you already know, and the same rate thinking runs the distance-and-speed problems in Part 2.

Learning goals

Turn a completion time into the rate $\tfrac{1}{T}$
Add or subtract work rates to find a team's time or a missing worker's time
Subtract a drain's rate from a filling rate, and tell whether the tank can fill at all
Apply $d = rt$ , solving for whichever is unknown
Compute average speed from totals, not from a mean

Part 1: Work rate problems

From time to rate

Suppose that a job is done at a steady pace, and that it takes $T$ units of time from start to finish. Then in one unit of time exactly the fraction $\frac{1}{T}$ of that job gets done. That fraction per unit time is the work rate. A pump that empties a pool in $5$ hours removes one fifth of the pool each hour, so its rate is $\frac{1}{5}$ pool per hour. In general,

\text{rate} = \frac{1}{T} \quad \text{(job per unit time)}.

Rate and time are reciprocals, the same inverse-variation relationship from the last lesson, now applied to work instead of speed. The amount of work is fixed at one whole job, so $\text{rate} \times \text{time} = 1$ , a constant product, which means the rate is inversely proportional to the time. Double the time and the rate halves; a faster rate goes with a shorter time. This reciprocal is the hinge of every work problem, so it is worth turning both ways:

\text{rate} = \frac{1}{\text{time}}, \qquad \text{time} = \frac{1}{\text{rate}}.

You almost always work in rates while solving, then flip the final rate back into a time at the very end.

Why work rates add

Rates combine cleanly in a way that times refuse to, and it is worth seeing exactly why before trusting it.

Worked example 1 Two painters, one room

One painter can paint a room alone in $4$ hours; a second painter can paint the same room alone in $6$ hours. Working together, how long do they take?

Turn each time into a rate. The first paints $\frac{1}{4}$ of the room per hour, the second paints $\frac{1}{6}$ per hour. Add the rates over the common denominator $12$ to get the combined rate:

\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \text{ of the room per hour}.

Together they paint $\frac{5}{12}$ of the room each hour. The time for the whole room is the reciprocal of that rate:

T = \frac{1}{\,5/12\,} = \frac{12}{5} = 2.4 \text{ hours}.

So they finish in $\frac{12}{5}$ hours, which is $2$ hours and $24$ minutes (since $0.4 \times 60 = 24$ ). Notice the answer is smaller than either painter’s solo time of $4$ or $6$ hours, exactly as it must be: two people working together beat either one alone.

In one hour, painter A finishes 1/4 of the job (3 of the 12 equal parts) and painter B finishes 1/6 (2 parts). Their rates add to 5/12 of the job per hour, so the team does more each hour than either painter alone.

Nothing about that reasoning depended on the specific numbers $4$ and $6$ . The same addition works for any two workers with steady rates on the same job, as long as they neither duplicate each other’s work nor slow each other down, which is worth stating in general.

Why work rates add#

Say one worker would finish the job alone in $T_1$ units of time and the other alone in $T_2$ units. Ask what happens in a single unit of time while both work at once. The first worker, going at the steady rate $\frac{1}{T_1}$ , completes $\frac{1}{T_1}$ of the job in that unit. The second, going at rate $\frac{1}{T_2}$ , completes $\frac{1}{T_2}$ of the job in the same unit. Assume the two do not get in each other’s way, and neither redoes work the other already finished. Then whatever the first finishes is simply work the second no longer has to do, so the fractions they complete add together. In that one unit of time the team finishes

\frac{1}{T_1} + \frac{1}{T_2}

of the job. But the fraction of a job that a team finishes in one unit of time is, by the definition above, exactly the team’s combined rate. So the combined rate is the sum of the individual rates,

\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2},

where $T$ is the time the team takes together. Rates add; times do not. This is why you can never average the two times, nor add them: it is the rates, the work done per unit of time, that combine. Only once you have added the rates do you take the reciprocal of that combined rate to get the time, provided that rate is positive. The argument extends without any change to three or more helpers, and even to helpers who work against each other. A drain that empties a tank is just a negative rate, and it enters the same sum with a minus sign.

Working together

As long as each worker holds a steady rate on the same job, and they neither slow each other down nor redo each other’s work, you find how long a team takes by adding the individual rates to get the combined rate, then taking the reciprocal of that sum, exactly as with the two painters above.

Check your understanding

One hose fills a pool in $3$ hours; another fills the same pool in $6$ hours. Running together, how long do they take to fill it?

4.5

hours B

9

hours C

2

hours D

3

hours

Finding an unknown time

The same equation runs in reverse. If you know the team’s time and every individual time but one, the missing rate is whatever remains after you subtract the known rates from the combined rate. Solve for that one rate, then flip it to get the time. Solving for the rate first sidesteps having to solve an equation with the unknown time buried inside a fraction.

Worked example 2 How long would the second worker take?

Two people raking leaves finish a yard together in $6$ hours. Working alone, the first would take $10$ hours. How long would the second take working alone?

Let the second worker’s solo time be $T_2$ hours, so the second rate is $\frac{1}{T_2}$ . The combined rate is $\frac{1}{6}$ per hour and the first rate is $\frac{1}{10}$ per hour, and the rates add:

\frac{1}{10} + \frac{1}{T_2} = \frac{1}{6}.

Solve for the unknown rate by subtracting the known one, using the common denominator $30$ :

\frac{1}{T_2} = \frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15}.

The second worker’s rate is $\frac{1}{15}$ of the yard per hour, so the solo time is the reciprocal:

T_2 = 15 \text{ hours}.

Check it: $\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$ , the combined rate, so the pair really does finish in $6$ hours.

Check your understanding

Two people can mow a lawn together in $4$ hours. Working alone, one of them would take $5$ hours. How long would the other take alone?

20

hours B

9

hours C

1

hour D

4.5

hours

Filling against a drain

A rate can be negative. A pipe that fills a tank adds a positive rate; a drain that empties it contributes a negative one, because it undoes work rather than doing it. The net rate is still just the signed sum of the separate rates. The time to fill the tank is the reciprocal of that net rate, provided the net rate is positive. If the drain were the faster of the two, the net rate would be negative and the tank would never fill at all. A $6$ -hour pipe against a $3$ -hour drain, for instance, nets $\frac{1}{6} - \frac{1}{3} = -\frac{1}{6}$ tank per hour: the negative sign means the tank loses ground every hour, so no matter how long you wait it never fills.

Worked example 3 A pipe and an open drain

A pipe fills an empty tank in $4$ hours. A drain at the bottom, left open, would empty a full tank in $6$ hours. If the pipe runs while the drain stays open, how long does the tank take to fill?

The pipe works at $+\frac{1}{4}$ tank per hour. The drain removes water, so it works at $-\frac{1}{6}$ tank per hour. Add the signed rates over the common denominator $12$ to get the net rate:

\frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \text{ tank per hour}.

The net rate is positive, so the tank does fill, gaining $\frac{1}{12}$ of a tank each hour. The time is the reciprocal of the net rate:

T = \frac{1}{\,1/12\,} = 12 \text{ hours}.

So the tank fills in $12$ hours. The open drain stretches a $4$ -hour fill out to $12$ hours, which fits the picture: some of every hour’s inflow is lost back out the bottom.

Check your understanding

A faucet fills a basin in $3$ hours. With the plug out, a drain would empty the full basin in $6$ hours. With the faucet on and the drain open, how long does the basin take to fill?

1.5

hours B

6

hours C

2

hours D

9

hours

Sometimes the drain wins. Check the net rate’s sign before you take a reciprocal of anything.

Check your understanding

A hose fills a tank in $8$ hours. A leak at the bottom would empty a full tank in $5$ hours. With the hose running and the leak open, what happens?

A The tank fills in

40

hours B The tank fills in

3

hours C The tank never fills; the net rate is negative D The tank never fills; the net rate is zero

Part 2: Distance and speed problems

The distance formula

This second family of rate problems trades “jobs per hour” for “miles per hour.” The formula has the same shape, distance equals rate times time, just as work equaled rate times time. The combining rule is different, though. Simultaneous workers who hold steady rates on the same job, without duplicating work or slowing each other down, combine by adding their rates; a multi-leg trip instead adds up the separate distances and the separate times, leg by leg, never the speeds themselves. Distance covered is rate multiplied by time:

d = rt.

Here $r$ is the speed, the distance covered per unit of time, and it plays the same role the work rate did. One relationship gives you all three questions, because you can solve it for whichever letter is unknown:

r = \frac{d}{t}, \qquad t = \frac{d}{r}, \qquad d = rt.

Worked example 4 A steady drive, two ways

A delivery van covers $165$ miles in $3$ hours at a steady speed. First find that speed, then find how far the van travels in $7$ hours at the same speed.

The speed is distance divided by time:

r = \frac{d}{t} = \frac{165}{3} = 55 \text{ miles per hour}.

Now hold that speed and run the relationship the other way, with $t = 7$ :

d = rt = 55 \times 7 = 385 \text{ miles}.

So the van travels at $55$ mph and covers $385$ miles in $7$ hours. The single relationship $d = rt$ answered both questions; you only changed which letter you solved for.

Check your understanding

A cyclist covers $18$ miles in $2$ hours at a steady speed. At that same speed, how long would it take to cover $63$ miles?

7

hours B

126

hours C

0.14

hours D

31.5

hours

Average speed done right

When a trip covers the same distance at each of two different speeds, its average speed is not the plain average of the two speeds: it always sits below that average. Average speed has one definition, and you must always go back to it, whatever the split:

\text{average speed} = \frac{\text{total distance}}{\text{total time}}.

Add up all the distance, add up all the time, and divide. Why the plain mean misses in this equal-distance case is worth pinning down, because the trap is so easy to fall into.

Why average speed is total distance over total time#

Average speed answers one question. If the whole trip had been run at a single steady speed, what speed would cover the same distance in the same total time? That is the speed you could replace the trip with and change nothing overall. So that speed must be the total distance divided by the total time, by the very definition $r = d/t$ applied to the trip as a whole.

Now see why, for equal distances at two different speeds, that plain average is too high. Because time is distance divided by speed, the slow leg takes more time than the fast leg: the slower you go, the longer you spend going. So when you form total distance over total time, the total time is weighted toward the slow leg, where you lingered. That weighting drags the average down toward the slower speed. The plain mean of the two speeds pretends you spent equal time at each, but you did not. You spent equal distance, and more time at the slower speed. That extra time at the low speed is exactly what pulls the true average below the halfway mark. The two averages coincide only when the two speeds are themselves equal, since equal distances at genuinely different speeds always eat unequal amounts of time.

Worked example 5 There and back at different speeds

You drive to a town $60$ miles away at $30$ mph, then return home along the same road at $60$ mph. What is your average speed for the whole round trip?

Resist averaging $30$ and $60$ to get $45$ . Go back to the definition, and build the total distance and the total time from the two legs.

The total distance is the round trip:

d = 60 + 60 = 120 \text{ miles}.

Now the time for each leg, using $t = d/r$ . The trip out at $30$ mph takes $\frac{60}{30} = 2$ hours; the trip back at $60$ mph takes $\frac{60}{60} = 1$ hour. So the total time is

t = 2 + 1 = 3 \text{ hours}.

Divide total distance by total time:

\text{average speed} = \frac{120}{3} = 40 \text{ miles per hour}.

The average is $40$ mph, below the naive $45$ mph. That is because you spent $2$ of the $3$ hours crawling at the slower $30$ mph and only $1$ hour at $60$ mph. More time at the low speed pulls the average down.

Check your understanding

A cyclist rides $12$ miles out at $12$ mph and returns the same $12$ miles at $4$ mph. What is the average speed for the whole ride?

8

mph B

6

mph C

5

mph D

9

mph

Common mistakes

Averaging or adding the times. Two workers’ solo times never average or add to give the team’s time. Add their rates instead, then take the reciprocal; for workers holding steady rates on the same job without duplicating work or slowing each other down, positive rates add to a rate exceeding either worker’s alone, so the team finishes faster than either could solo.
Forgetting to flip the rate back to a time. Adding rates gives you a combined rate, not a time. The question asks for a time, so the last step is a reciprocal, and that reciprocal is a completion time only when the combined or net rate is positive; a zero or negative net rate means the job never finishes at all.
Adding a drain instead of subtracting it. A drain undoes work, so its rate enters the sum with a minus sign, never a plus sign.
Averaging the two speeds. Average speed is total distance over total time, not simply the mean of the speeds. The round trip above shows why: the true average sits below the mean whenever the two speeds differ, and equals it only when the two speeds are the same.
Mixing up the parts of $d = rt$ . Keep the units straight: a speed is miles per hour, a time is hours, and their product is miles. Divide to get a time, multiply to get a distance.

Key takeaways

A job finished in time $T$ is done at the rate $\frac{1}{T}$ per unit time, and rate and time are reciprocals: $\text{time} = \frac{1}{\text{rate}}$ .
Rates add when workers hold steady rates on the same job without duplicating work or slowing each other down: $\frac{1}{T_1} + \frac{1}{T_2} = \frac{1}{T}$ . Add the rates, then take the reciprocal to get the together time $T$ .
To find a missing worker’s time, subtract the known rates from the combined rate, then flip the leftover rate.
A drain is a negative rate: the net rate is the signed sum, and the fill time is its reciprocal when the sum is positive.
Distance, rate, time: $d = rt$ , so $r = \frac{d}{t}$ and $t = \frac{d}{r}$ ; solve for whichever one is unknown.
Average speed is $\dfrac{\text{total distance}}{\text{total time}}$ , not the mean of the speeds. Over equal distances at two different speeds it falls below the mean, because more time is spent at the slower speed.

Rate and Work Problems

Learning goals

Part 1: Work rate problems

From time to rate

Why work rates add

Working together

Finding an unknown time

Filling against a drain

Part 2: Distance and speed problems

The distance formula

Average speed done right

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice